Exploring the Definition of <u|A|u>

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In summary, there are two different definitions for <u|A|u>, one involving the length of the vector A|u> squared and the other involving the mean/average of the observable A on the quantum state |u>. These definitions are not equal, as they involve different elements. The selfadjoint linear operator A=|n><n| is an example of an orthogonal projector, where repeated projections do not change anything.
  • #1
trosten
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How do u define this <u|A|u>? where A is a selfadjoint linear operator.

I have one book that defines is as ||A|u>||^2=<u|A|u>
(equation 2.6 http://www.theory.caltech.edu/people/preskill/ph229/notes/chap2.ps)

and since ||A|u>||^2 is the length of the vector A|u> squared that is equal to (<u|A)(A|u>).

I have another book (j.j sakurai) that defines it as (<u|)(A|u>)=(<u|A)(|u>)=<u|A|u> but this is the projection of A|u> on |u>.

It seems to me that the two definitions arent equal !? Any ideas? :confused:
 
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  • #2
Of course,they're not equal,as they involve different "animals".

[tex] \langle \psi|\hat{A}|\psi\rangle [/tex]

is the mean/avrerage of the observable A described by the selfadjoint operator [itex] \hat{A} [/itex] on the quantum state [itex] |\psi \rangle [/itex]
,while the norm
[tex] || \hat{A}|\psi\rangle ||^{2} [/tex]

involves the matrix element of the SQUARE of the operator...

Daniel.

P.S.The operator is selfadjoint,that's why is the square...
 
Last edited:
  • #3
I found the solution to the problem myself. If the operator is A=|n><n| then ofcourse AA = (|n><n|)(|n><n|) = |n><n|n><n| = |n><n| = A
Ofcourse repeated projections shouldn't change anything! silly me.
 
  • #4
OKay,u should have told right from the beginning that u're interested in true and IDEMPOTENT ortogonal projectors... :wink:

Daniel.
 
  • #5
haha yeah didnt think about that :smile:
 

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